To find the function f given the second derivative f''(x) and the boundary conditions f(0) = 0 and f(π) = 0, we can follow these steps:
1. Start by integrating the second derivative f''(x) with respect to x to find the first derivative f'(x). Keep in mind that when you integrate, you need to add a constant of integration (C).
∫ (f''(x)) dx = ∫ (3e^x + 3sin(x)) dx
= 3∫ (e^x) dx + 3∫ (sin(x)) dx
= 3(e^x) - 3(cos(x)) + C1
2. Now, integrate the first derivative f'(x) to find the original function f(x). Again, include a constant of integration (D).
∫ (f'(x)) dx = ∫ (3e^x - 3cos(x) + C1) dx
= 3∫ (e^x) dx - 3∫ (cos(x)) dx + ∫ (C1) dx
= 3(e^x) - 3(sin(x)) + C1x + D
3. Apply the boundary conditions f(0) = 0 and f(π) = 0 to determine the values of the constants C1 and D.
For f(0)=0:
f(0) = 3e^0 - 3sin(0) + C1(0) + D
0 = 3 - 0 + 0 + D
D = -3
For f(π)=0:
f(π) = 3e^π - 3sin(π) + C1(π) - 3
0 = 3e^π - 0 + C1π - 3
C1 = (3 - 3e^π)/π
4. Substitute the values of C1 and D back into the equation for f(x) to get the final result:
f(x) = 3e^x - 3sin(x) + ((3 - 3e^π)/π) x - 3